Moments are popularly used to describe the characteristic of a distribution. The Greek letter \(\mu\) (read as mu) is generally used to denote moments.
Suppose a variable \(x\) changes values \(x_1, x_2, x_3, \ldots, x_m\) then:
\[\bar{X}^m = \frac{x^n + x_1^n + \ldots + x_n^n}{m} = \bar{X}^m\]
\[\frac{\sum_{i=1}^{n} x_i^n}{m}\]
is called the \(n^{th}\) moment about the origin (raw moment).
Example 4.1
\(X = 10, 11, 12\)
\(\bar{X}^2\) is called \(2^{nd}\) moment about the origin:
\[\bar{X}^2 = \frac{10^2 + 11^2 + 12^2}{3}\]
\(\bar{X}\) is called the first moment about the origin (mean, \(E(X)\)):
\[\bar{X} = \frac{10 + 11 + 12}{3}\]
Given a data set \(x_1, x_2, x_3, \ldots, x_n\), the \(r^{th}\) moment of a variable \(X\) about the arithmetic mean is given by:
\[M_r = \frac{\sum_{i=1}^{n}(x_i - \bar{X})^r}{n} \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text{(i)} \quad \text{(ungrouped data)}\]
Example 4.2
\(X = 10, 11, 12\)
First moment about mean:
\[\bar{X} = 11\tfrac{2}{3}\]
\[\frac{\left(10 - 11\tfrac{2}{3}\right) + \left(12 - 11\tfrac{2}{3}\right) + \left(13 - 11\tfrac{2}{3}\right)}{3} = 0\]
N/B: The first moment about the mean is always 0.
Proof: \(1^{st}\) moment about mean is zero
\[M_r = \frac{\sum_{i=1}^{m} x_i - \bar{X}^r}{m}\]
\[= \frac{\sum_{i=1}^{m} x_i}{m} - \frac{\sum_{i=1}^{m} \bar{X}}{m}\]
\[= \bar{X} - \bar{X} = 0\]
Second moment about the mean:
\[M_2 = \frac{\sum_{i=1}^{m}(x_i - \bar{X})^2}{m} = S^2 \quad \text{or variance}\]
The \(r^{th}\) moment of a variable \(X\) about any arbitrary point \(A\) is given by:
\[M_r' = \frac{\sum_{i=1}^{m}(x_i - a)^r}{m} \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text{(ii)}\]
Example 4.3
\(X = 10, 12, 13\), \(a = 10\) (First moment about 10):
\[M_r' = \frac{(10-10) + (12-10) + (13-10)}{3}\]
\[= \frac{2 + 3}{3} = \frac{5}{3}\]
Remarks
\(E(X^r)\) — \(r^{th}\) moment about the origin:
\[\frac{\sum x^r}{m}\]
\(r^{th}\) Moment about origin:
\[M_1 = 0\] \[M_2 = S^2\] \[\vdots\] \[M_r = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^r}{n}\]
Moments about any point \(a\):
\[M_r' = \frac{\sum_{i=1}^{n}(x_i - a)^r}{n}\]
Example 4.4
Find the first, second, third & fourth moments of the following set of numbers about the origin:
\(2, 3, 7, 8, 10\)
\[M_1 = \bar{X} = \frac{2+3+7+8+10}{5} = 6\]
\[M_2 = \bar{X}^2 = \frac{2^2+3^2+7^2+8^2+10^2}{5} = 45.2\]
\[M_3 = \bar{X}^3 = \frac{2^3+3^3+7^3+8^3+10^3}{5} = 378\]
\[M_4 = \bar{X}^4 = \frac{2^4+3^4+7^4+8^4+10^4}{5} = 3318.8\]
Prove that: \(M_2 = M_2' - [M_1']^2\)
Let \(d = x - a\), so \(x = a + d\), \(\bar{X} = a + \bar{d}\), and \(x - \bar{X} = d - \bar{d}\).
\[M_2 = \frac{\sum(x - \bar{X})^2}{n}\]
\[= \frac{\sum(d - \bar{d})^2}{n} = \frac{\sum(d^2 - 2d\bar{d} + \bar{d}^2)}{n}\]
\[= \frac{\sum d^2}{n} - 2\bar{d}\left(\frac{\sum d}{n}\right) + \frac{\sum(\bar{d})^2}{n}\]
\[= \frac{\sum d^2}{n} - 2\bar{d}\left(\frac{\sum d}{n}\right) + (\bar{d})^2\]
\[= \frac{\sum d^2}{n} - 2\bar{d}\cdot\bar{d} + (\bar{d})^2\]
\[= \frac{\sum d^2}{n} - (\bar{d})^2\]
\[= \left(\frac{\sum(x-a)}{n}\right)^2 - \left(\frac{\sum(x-a)}{n}\right)^2\]
\[= M_2' - \left(\frac{\sum d}{n}\right)^2 = M_2' - \left(\frac{\sum(x-a)}{n}\right)^2\]
\[\boxed{M_2 = M_2' - [M_1']^2}\]
Prove that: \(M_3 = M_3' - 3M_1'M_2' + 2(M_1')^3\)
\[M_3 = \frac{\sum(x - \bar{X})^3}{n} = \frac{\sum(d - \bar{d})^3}{n}\]
\[= \frac{\sum\left(d^3 - 3d^2\bar{d} + 3d(\bar{d})^2 - (\bar{d})^3\right)}{n}\]
\[= \frac{\sum d^3}{n} - 3\bar{d}\left(\frac{\sum d^2}{n}\right) + 3(\bar{d})^2\frac{\sum d}{n} - (\bar{d})^3\]
\[= \frac{\sum d^3}{n} - 3\bar{d}\left(\frac{\sum d^2}{n}\right) + 3(\bar{d})^2\bar{d} - (\bar{d})^3\]
\[= \frac{\sum d^3}{n} - 3\bar{d}\left(\frac{\sum d^2}{n}\right) + 2(\bar{d})^3\]
\[= \bar{d}^3 - 3\bar{d}\cdot\bar{d}^2 + 2(\bar{d})^3\]
\[= \left(\frac{\sum(x-a)}{n}\right)^3 - 3\left(\frac{\sum(x-a)}{n}\right)\left(\frac{\sum(x-a)^2}{n}\right) + 2\left(\frac{\sum(x-a)}{n}\right)^3\]
\[\boxed{M_3 = M_3' - 3M_1'M_2' + 2(M_1')^3}\]
Prove that: \(M_4 = M_4' - 4M_1'M_3' + 6(M_1')^2 M_2' - 3(M_1')^4\)
\[M_4 = \frac{\sum(x - \bar{x})^4}{n} = \frac{\sum(d - \bar{d})^4}{n}\]
\[= \frac{\sum\left(d^4 - 4d^3(\bar{d}) + 6d^2(\bar{d})^2 - 4d(\bar{d})^3 + (\bar{d})^4\right)}{n}\]
\[= \bar{d}^4 - 4\bar{d}\cdot\bar{d}^3 + 6(\bar{d})^2\cdot\bar{d} - 4(\bar{d})^3\cdot\bar{d} + (\bar{d})^4\]
\[= \bar{d}^4 - 4\bar{d}\cdot\bar{d}^3 + 6(\bar{d})^2\cdot\bar{d} - 4(\bar{d})^4 + (\bar{d})^4\]
\[= \bar{d}^4 - 4\bar{d}\cdot\bar{d}^3 + 6(\bar{d})^2\cdot\bar{d} - 3(\bar{d})^4\]
\[= \left(\frac{\sum(x-a)}{n}\right)^4 - 4\left(\frac{\sum(x-a)}{n}\right)\left(\frac{\sum(x-a)^3}{n}\right) + 6\left(\frac{\sum(x-a)}{n}\right)^2\left(\frac{\sum(x-a)^2}{n}\right) - 3\left(\frac{\sum(x-a)}{n}\right)^4\]
\[\boxed{M_4 = M_4' - 4M_1'M_3' + 6(M_1')^2 M_2' - 3(M_1')^4}\]
For Grouped Data
The \(r^{th}\) moment about the origin is given by:
\[\bar{X}^r = \frac{f_1 x_1^r + f_2 x_2^r + \ldots + f_n x_n^r}{f_1 + f_2 + \ldots + f_n} = \frac{\sum_{i=1}^{n} f_i x_i^r}{\sum f_i}\]
\(r^{th}\) Moment about the mean:
\[M_r = \frac{\sum_{i=1}^{n} f_i (x_i - \bar{X})^r}{\sum f_i}\]
For different values of \(r\), we shall get different moments. Thus if we put \(r = 1\), we will get the first moment, if we put \(r = 2\), we will get the second moment, and so on.
\[M_1 = 0 \qquad M_2 = S^2\]
\(r^{th}\) Moment about any point \(a\):
\[M_r' = \frac{\sum_{i=1}^{n} f_i(x_i - a)^r}{\sum f_i}\]
If \(X_i = A + CU_i\) where \(A\) and \(C\) are constants, then:
\[M_r' = \frac{C^r \sum f_i u_i}{\sum f_i}\]
Example 4.5
The marks obtained by students in a class were as follows:
| Mark | Frequency | X | D | U = d/c, c = 10 | fu | fu² | fu³ | fu⁴ |
|---|---|---|---|---|---|---|---|---|
| 1-10 | 4 | 5.5 | -40 | -4 | -16 | 64 | -256 | 1024 |
| 11-20 | 5 | 15.5 | -30 | -3 | -15 | 45 | -135 | 405 |
| 21-30 | 32 | 25.5 | -20 | -2 | -64 | 128 | -256 | 512 |
| 31-40 | 89 | 35.5 | -10 | -1 | -89 | 89 | -89 | 89 |
| 41-50 | 102 | 45.5 | 0 | 0 | 0 | 0 | 0 | 0 |
| 51-60 | 78 | 55.5 | 10 | 1 | 78 | 78 | 78 | 78 |
| 61-70 | 63 | 65.5 | 20 | 2 | 126 | 252 | 504 | 1008 |
| 71-80 | 21 | 75.5 | 30 | 3 | 63 | 189 | 567 | 1701 |
| 81-90 | 9 | 85.5 | 40 | 4 | 36 | 144 | 576 | 2304 |
| 91-100 | 3 | 95.5 | 50 | 5 | 15 | 75 | 375 | 1875 |
| Total | 406 | 134 | 134 | 1064 | 1364 | 8996 |
Obtain the first 4 moments about the mean.
Moments about 45.5:
\[M_1' = \frac{C\sum fu}{\sum f} = \frac{10 \times 134}{406} = 3.30\]
\[M_2' = \frac{C^2\sum fu^2}{\sum f} = \frac{100 \times 1064}{406} = 262.07\]
\[M_3' = \frac{C^3\sum fu^3}{\sum f} = \frac{1000 \times 1364}{406} = 3359.61\]
\[M_4' = \frac{C^4\sum fu^4}{\sum f} = \frac{10000 \times 8996}{406} = 221576.35\]
\[M_1 = 0\]
\[M_2 = M_2' - [M_1']^2 = 262.07 - (3.30)^2 = 251.18\]
\[M_3 = M_3' - 3M_1'M_2' + 2(M_1')^3 = 3359.61 - 3(3.30)(262.07) + 2(3.30)^2 = 786.90\]
\[M_4 = M_4' - 4M_1'M_3' + 6(M_1')^2 M_2' - 3(M_1')^4\] \[= 221576.35 - 4(3.30)(3359.35) + 6(3.3)^2(262.07) - 3(3.30)^4 = 194353.15\]
Example 4.6
Calculate the first four moments about the mean of the following distribution:
| x | f | d (x - a) | u | fu | fu² | fu³ | fu⁴ |
|---|---|---|---|---|---|---|---|
| 12 | 1 | -6 | -3 | -3 | 9 | -27 | 81 |
| 14 | 4 | -4 | -2 | -8 | 16 | -32 | 64 |
| 16 | 6 | -2 | -1 | -6 | 6 | -6 | 6 |
| 18 | 10 | 0 | 0 | 0 | 0 | 0 | 0 |
| 20 | 7 | 2 | 1 | 7 | 7 | 7 | 7 |
| 22 | 2 | 4 | 2 | 4 | 8 | 16 | 32 |
| Total | 30 | -6 | 46 | -42 | 190 |
\[M_1' = \frac{C\sum fu}{\sum f} = \frac{2 \times (-6)}{30} = -0.4\]
\[M_2' = \frac{C^2\sum fu^2}{\sum f} = \frac{4 \times 46}{30} = 6.13\]
\[M_3' = \frac{C^3\sum fu^3}{\sum f} = \frac{8 \times (-42)}{30} = -11.2\]
\[M_4' = \frac{C^4\sum fu^4}{\sum f} = \frac{16 \times 190}{30} = 101.33\]
\[M_1 = 0\]
\[M_2 = M_2' - [M_1']^2 = 6.13 - (-0.4)^2 = 5.97\]
\[M_3 = M_3' - 3M_1'M_2' + 2(M_1')^3 = (-11.2) - 3(-0.4)(6.13) + 2(-0.4)^3 = -3.524\]
\[M_4 = M_4' - 4M_1'M_3' + 6(M_1')^2 M_2' - 3(M_1')^4\] \[= 101.33 - 4(-0.4)(-11.2) + 6(-0.4)^2(6.13) - 3(-0.4)^4 = 89.294\]